Which integral gives the length of the arc of the equation $y=\dfrac1x$ from the point $\left(\dfrac{1}{2},2\right)$ to the point $\left(2,\dfrac12\right)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{1/2}^2 \sqrt{1-\dfrac1{x^2}}\,dx$ (Choice B) B $ \int_{1/2}^2 \sqrt{1+\dfrac1{x^2}}\,dx$ (Choice C) C $ \int_{1/2}^2 \sqrt{1+\dfrac1x}\,dx$ (Choice D) D $ \int_{1/2}^2 \sqrt{1+\dfrac1{x^4}}\,dx$
The arc length $L$ of the curve $y=f(x)$ over the interval $[a, b]$ is $ L=\int_a^b\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\, dx$. First, calculate $dy/dx$. $\dfrac{dy}{dx}=-\dfrac1{x^2}$ Finally, apply the arc length formula on the interval $[1/2,\,2]$ and simplify the integral. $\begin{aligned} L &= \int_{1/2}^2 \sqrt{1+\left(-\dfrac1{x^2}\right)^2}\,dx \\\\ &= \int_{1/2}^2 \sqrt{1+\dfrac1{x^4}}\,dx \end{aligned}$